Hi all,
I'm new to FFT and I want to ask about calculating FFT in Matlab. The following is my coding guidance from Matlab help:-
clc;
Fs = 512; % Sampling frequency
T = 1/Fs; % Sample time
L = 8100; % Length of signal
t = (0:L-1)*T; % Time vector
% Sum of a 50 Hz sinusoid and a 120 Hz sinusoid
x = 0.7*sin(2*pi*50*t) + sin(2*pi*120*t);
y = x + 2*randn(size(t)); % Sinusoids plus noise
plot(Fs*t(1:50),y(1:50));
title('Signal Corrupted with Zero-Mean Random Noise')
xlabel('time (milliseconds)');
NFFT = 2^nextpow2(L); % Next power of 2 from length of y
Y = fft(y,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2);
% Plot single-sided amplitude spectrum.
plot(f,2*abs(Y(1:NFFT/2)));
title('Single-Sided Amplitude Spectrum of y(t)');
xlabel('Frequency (Hz)');
ylabel('|Y(f)|');
Can someone verify my code because the fft graph unconvincing. Here the figure of the graph
Thank you.

댓글 수: 5

Walter Roberson
Walter Roberson 2015년 6월 19일
What features of the plot are you seeing as "unconvincing" ?
nur yusof
nur yusof 2015년 6월 29일
the peak.looks not relevant.
Walter Roberson
Walter Roberson 2015년 6월 29일
I am not sure why you say that? You coded
0.7*sin(2*pi*50*t) + sin(2*pi*120*t);
which calls for a 0.7 peak at 50 Hz and a 1.0 peak at 120 Hz. The peaks are visually shown at the correct place, 50 Hz and 120 Hz. When I extract the value of the peak locations I get 0.683776526075202 and 0.986050681728091 . Why not exactly 0.7 and 1.0? Because of the normally distributed random noise with standard deviation 2 that you added, a standard deviation that exceeds the maximum value of the two sines added together (0.7+1). The peaks look relevant to me.
Sag
Sag 2016년 1월 26일
Why to multiply by 2 in step: plot(f,2*abs(Y(1:NFFT/2)));
Afshin Loghmani M Toussi
Afshin Loghmani M Toussi 2020년 12월 1일
Because you are plotting half of the frequency range( because of the symmetry). However, to show the same amount of energy you are multiplying the values by 2.

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Andreas Goser
Andreas Goser 2015년 6월 19일

0 개 추천

Frequent puzzlese with interpreting FFT:
2 peaks instead of 1 peak
Frequency "off" by a factor of 6.
Both effects are in the nature of the algorithm and "6" is actually "2*Pi". Next steps depend on whether you want to just use the results, then this will help. If you want to understand, read a text book about signal processing.

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2015년 6월 19일

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2020년 12월 1일

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